This problem is really weird. I have to explain why MVT applies for f(x)=2sinx-sin2x on the closed interval 7pi,8pi and then determine all values of c in the interval (7pi,8pi) that satisfies the conclusion of the theorem. However, the condition of MVT is that the function is continous, but the domain of the function is (-pi, 3pi), technically implicating that the function in discontinous on the interval (7pi,8pi). So how would I solve this problem?
why do you think the domain is from -pi to 3pi ? I think it is all real x
df/dx = 2 cos x -2 cos 2x (hoping you did not mean sin^2x)
at 7 pi
2sin 7pi = 0
in fact sin any number of pis is zero
so we are looking for where the slope is 0
that is where cos x = cos2x
well, that is true when cos x = 0
which is pi/2 + 2 pi n and 3 pi/2 + 2 pi n
between
7 pi and 8 pi that is
14 pi/2 + pi/2 = 15 pi/2
14pi/2 + 3pi/2 = 17 pi/2
I see. I thought it was -pi to 3pi because the period was 2pi when I was inputting the domain and range.
So to reiterate, the values of c that satisfy the conclusion of the theorem are 15pi/2 and 17pi/2?
f(x)=2sinx-sin2x
f(7pi) = 0
f(8pi) = 0
f(x) is continuous.
So, it satisfies the MVT. In fact, it satisfies Rolle's Theorem.
(f(8pi)-f(7pi))/(8pi-7pi) = 0
f'(x) = 2cosx - 2cos(2x)
= 2cosx - 2(2cos^2x-1)
= -4cos^2x + 2cosx + 2
= -2(2cos^2x - cosx - 1)
= -2(2cosx+1)(cosx-1)
So, you want c such that f'(c) = 0
cosx = -1/2: x = pi/3+7pi = 22pi/3
or
cosx = 1: x = 8pi
We want x in the interval (7pi,8pi), so that means x = 22pi/3 ? 23.03
http://www.wolframalpha.com/input/?i=2sinx-sin2x,+7pi+%3C%3D+x+%3C%3D+8pi
cos x =cos 2x
cos 2x = cos^2x -sin^2 x
where does cos^2x -(1-cos^2x) =cos x cos x = 2 cos^2x-1
2 cos^2x -cos x - 1 = 0
(2 cos x +1)(cos x -1) = 0
cos x = 1 or cos x = -1/2
x = 0 or 2 pi n
or x = 2 pi/3 +2 pi n
To explain why the Mean Value Theorem (MVT) applies for the function f(x) = 2sin(x) - sin(2x) on the closed interval [7π, 8π], and determine the values of c that satisfy the conclusion of the theorem, we need to address the issue of continuity.
While the domain of the function f(x) is (-π, 3π), it doesn't necessarily imply that the function is discontinuous on the interval (7π, 8π). The function might extend beyond its domain and still remain continuous on the closed interval [7π, 8π]. Therefore, it's important to examine the behavior of the function f(x) in that specific interval.
To determine if the function is continuous on the interval (7π, 8π), you need to check three conditions:
1. The function is defined at every point in the interval.
2. The limit from the left and the limit from the right exist and are equal at every point in the interval.
3. The function value at each endpoint of the interval is equal to the limit from the left and the limit from the right.
If all three conditions are satisfied, the function is continuous on the interval.
Once you determine that the function is indeed continuous on (7π, 8π), you can apply the Mean Value Theorem, which states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) where the derivative of the function is equal to the average rate of change of the function on the interval [a, b].
In other words, for the function f(x) on the closed interval [7π, 8π], there exists at least one value c in (7π, 8π) such that f'(c) = (f(8π) - f(7π))/(8π - 7π).
To determine the values of c that satisfy the conclusion of the Mean Value Theorem, you can follow these steps:
1. Find the derivative of f(x) = 2sin(x) - sin(2x).
2. Evaluate the function at the endpoints of the interval, f(8π) and f(7π), and calculate the difference f(8π) - f(7π).
3. Calculate the difference between the endpoints of the interval, (8π - 7π).
4. Divide the difference in function values by the difference in x-values to find the average rate of change.
5. Set the derivative of f(x) equal to the average rate of change and solve for c.
By solving this equation, you can find all the values of c in the interval (7π, 8π) that satisfy the conclusion of the Mean Value Theorem.